If the sum of infinite terms of a G.P. is 3 a | vedclass

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(A) Let the first term be $a$ and the common ratio be $r$. The sum of infinite terms of a $G.P.$ is given by $S = \frac{a}{1-r} = 3$ .....$(i)$The ter

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$3/2, 1/2$

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(A) Let the first term be $a$ and the common ratio be $r$. The sum of infinite terms of a $G.P.$ is given by $S = \frac{a}{1-r} = 3$ .....$(i)$The terms of the $G.P.$ are $a, ar, ar^2, \dots$. The squares of these terms are $a^2, a^2r^2, a^2r^4, \dots$,which form a new $G.P.$ with first term $a^2$ and common ratio $r^2$.The sum of the squares of the terms is $\frac{a^2}{1-r^2} = 3$ .....(ii)From $(i)$,$a = 3(1-r)$. Substituting this into (ii):$\frac{[3(1-r)]^2}{1-r^2} = 3$$\frac{9(1-r)^2}{(1-r)(1+r)} = 3$$\frac{3(1-r)}{1+r} = 1$$3 - 3r = 1 + r$$4r = 2 \Rightarrow r = 1/2$Substituting $r = 1/2$ into $(i)$:$a = 3(1 - 1/2) = 3(1/2) = 3/2$Thus,the first term is $3/2$ and the common ratio is $1/2$.

If the sum of infinite terms of a $G.P.$ is $3$ and the sum of the squares of its terms is $3$,then its first term and common ratio are

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